Download OX(D) (or O(D)) for a Cartier divisor D on a scheme X (1) on

The notions of OX (D) for an eective Cartier divisor D on a scheme X and OPnk (1) on projective
space have some dierent features, but on projective space OPnk (H) (for a hyperplane H ) coincides
with OPnk (1), so it may get confusing.
I'll describe the 2 seperately, and then see how it goes to the same thing on Pn .
(Remarks : Hartshorne may give more details than Voisin if you want very accurate denitions. In
Hartshorne OX (D) is denoted L (D). Also note that OX (D) is dened for arbitrary Cartier divisors,
not only the eective ones. We do not care about noneective divisors here.)
OX (D) (or O(D)) for a Cartier divisor D on a scheme X
Recall that an eective Cartier divisor D is dened, locally on some open ane U = Spec(A), by
a nonzerodivisor equation f ∈ A up to units in A (cf Hartshorne p. 141 and 145). To it, is associated
the invertible sheaf OX (D) dened on U as the module f1 A.
Remark 1 : in this general denition of OX (D), there is NO notion of degree involved.
Remark 2 : there is an injective map OX → OX (D), which locally is the map A → f1 A, a 7→ a = f1 f a.
In our example X = Pnk and the divisor D is a hyperplane H with equation a linear form `.
(Let us not assume that ` = xn , since it is actually useless.) Let's look at the open ane Ui =
{xi 6= 0}, its function ring is Ai = k[ xx0i , ..., xxni ], the ring of degree 0 elements inside the localization
Bi = k[x0 , ..., xn , x1i ]. (Note : Ai is not graded.) Let `i = `/xi be the image of ` in Ai . Then on Ui the
sheaf OPnk (H) is `1i Ai and we have the map Ai → `1i Ai dened above.
OPnk (1) on projective space
The sheaf OPnk (1) on X = Pnk is dened as the sheaf of rational functions inside K that have degree 1
(K is the constant sheaf on X dened by the fraction eld K of B = k[x0 , ..., xn ]). More precisely, on
Ui it is the sheaf associated to the module Mi of degree 1 elements inside K = Frac(Bi ). Obviously
Mi = xi Ai , since xi is invertible in Bi .
The isomorphisms ϕij : (OPnk (1)|Ui )|Uj → (OPnk (1)|Uj )|Ui are just the identity mappings. More
precisely, the function ring of Uij = Ui ∩ Uj may be seen by localizing Ai at xxji , we get Aij =
x
k[ xx0i , ..., xxni , xxji ], or by localizing Aj at xxji , we get Aji = k[ xx0j , ..., xxnj , xji ] which is of course the same
thing. Then ϕij above is the identity map xi Aij → xj Aji = xj Aij :
xi a 7→ xi a = xj
xi
a.
xj
The construction of O(1) actually makes sense starting from any graded ring (so one denes O(1)
not just for projective space), and what's essential is the notion of degree. But in full generality, I
think that there is no map OX → O(1). See the dierences with OX (D) above.
OPnk (H) ' OPnk (1) on projective space
Let X = Pnk again. We dene an isomorphism OPnk (H) ' OPnk (1) which is just multiplication by `.
On the open ane Ui , this translates as multiplication by `i xi since ` = `i xi :
1
`i x i
Ai −→
xi Ai
`i
On Ui we have the following commutative square : the top row is the map OPnk → OPnk (H), the right
vertical arrow is the isomorphism OPnk (H) ' OPnk (1), and the bottom row is the map that comes as a
consequence of this isomorphism :
Ai Ai

/ 1 Ai
`i
`i x i
`i x i
/ xi Ai
So we have a morphism OPnk → OPnk (1), but we see that it owes its existence to the hyperplane H ...
In other words there is no canonical morphism OPnk → OPnk (1), we have to choose a hyperplane. (This
is consistent with the fact that there is no map OX → O(1) in general.)